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The question is:

For a>0 and $\sqrt{a}+\frac{1}{\sqrt{a}}=3$, find the value of $a\sqrt{a}+\frac{1}{a\sqrt{a}}$

So I first squared the given equation and got:

$$a+\frac{1}{a}+2=9$$ $$a+\frac{1}{a}=7$$

Then to get the form of $a\sqrt{a}+\frac{1}{a\sqrt{a}}$:

$$(a+\frac{1}{a})(\sqrt{a}+\frac{1}{\sqrt{a}})=a\sqrt{a}+\frac{1}{a\sqrt{a}}+\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}$$

And I just got stuck right here because I didn't really know what to do with the $\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}$. So I looked into the solutions and apparently it's

$$(a+\frac{1}{a})(\sqrt{a}+\frac{1}{\sqrt{a}})=a\sqrt{a}+\frac{1}{a\sqrt{a}}+\sqrt{a}+\frac{1}{\sqrt{a}}$$

I'm not sure how those two are equal...

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Note that $$ \frac{\sqrt a}a = \frac{\sqrt a}{\sqrt a\cdot\sqrt a} = \frac1{\sqrt a}\\ \frac a{\sqrt a} = \frac{\sqrt a\cdot \sqrt a}{\sqrt a} = \sqrt a $$ From this we get that "those two" are indeed equal.

The top equality might be worth remembering specifically, because some people prefer to write $\frac1{\sqrt 2}$, while some people prefer to write $\frac{\sqrt2}2$. Knowing that it's the same number, without having to check every time, will make reading them faster.

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Let $x=\sqrt{a}$ then $x+\frac{1}{x}=3$ and $$ \\a\sqrt{a}+\frac{1}{a\sqrt{a}}=x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3x\cdot\frac{1}{x}(x+\frac{1}{x})=27-3\cdot3=18 $$

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$\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}=\sqrt{a}+\frac{1}{\sqrt{a}}=3$ !

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  • $\begingroup$ $3!$ is not the right answer (be careful with exclamation marks). $\endgroup$ – Arthur Nov 7 '18 at 12:41

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